# Tag Info

0

The difference lies in the dimension of the Hilbert space generated. Take $\{ \xi_1, \xi_2\} \sim N(0, I_2)$, which form an orthonormal basis of a two dimensional subspace. This is clearly not the same as the one dimensional subspace generated by $\xi_1$. Indeed you seem to be missing something fundamental: the distribution of a random variable $\xi$, or ...

3

Since $A^2$ is compact, it has an eigenvalue $\alpha$. As $A^2$ is self-adjoint, $\alpha$ is real. Let $\lambda\in\mathbb C$ with $\lambda^2=\alpha$. As $\alpha$ is an eigenvalue for $A^2$, we have that there exists nonzero $v$ with $(A^2-\alpha I)v=0$. Then $$0=(A^2-\alpha I)v=(A-\lambda I)(A+\lambda I)v=0.$$ If $(A+\lambda I)v=0$, then $-\lambda$ is ...

3

For this to hold it is crucial that $H$ is a complex Hilbert space since on $\mathbb{R}^2$ a rotation $T$ by $\pi/2$ has no eigenvalues, yet $T^2 = -I$ is compact and self-adjoint. The spectral theorem for compact self-adjoint operators yields an eigenvector $w$ of $A^2$ with real eigenvalue $\lambda$. Without loss of generality we can rescale $A$ ...

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This is the first thought that I had. I do not see how it leads to what you want but maybe you have an idea with it: Since $A^2$ is a self-adjoint compact operator either $||A||^2$ or $-||A||^2$ is an eigenvalue for $A^2$.

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Since $P+Q=(P+Q)^2$, we get $P+Q=P+Q+PQ+QP$, so $PQ+QP=0$. Multiplying by $P$ both left and right, we get $2PQP=0$, so $PQP=0$. Now we use that $P^*=P$, $Q^*=Q$: $$0=PQP=PQQP=P^*Q^*QP=(QP)^*QP,$$ so $QP=0$. Then $0=\langle QPx,y\rangle=\langle Px,Qy\rangle$ for any $x,y$, showing that the ranges are orthogonal.

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It seems the following. Proposition. Let $V$ be a separable real Hilbert space, $V’$ be a normed space and $f:V\to V’$ be a continuous map. Then there exists a sequence $\{f_n\}$ of bounded Lipschitz maps from $V$ to $V’$ converging uniformly to $f$ on compact sets. Proof. It suffices to consider the case $V=\ell_2$. Let $n$ be a natural number. We shall ...

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Thanks to those who replied. I have gathered bits from the answers and decided to write my own answer to better understand the solutions. 1) Fix $h\in H$. First, I claim that there can only be countably many $v_k$ in our orthonormal basis set $S$ for which $\langle v_k,h\rangle \neq0$. Proof: Consider the following partition of the non-negative reals: ${0}, ... 0 I have been struggling with the same questions when I was studying Hilbert Spaces first time. But I found following answers. 1) every vector in the Hilbert space is NOT a (infinite) combination of the basis vectors (I mean, what does sum of infinite number of "vectors" mean in the first place?). What we have is an isometrical isomorphism between the Hilbert ... 0 For the first one, the answer is yes. In metric spaces (more generally, first countable spaces), in particular in Hilbert spaces, the closure of a set is just the set of all limits of sequences. As you have said, the set of finite combinations of basic vectors is dense, so any element of the space is limit of finite combinations; and it's not hard to see ... 0 1) The sum of over an uncountable set of indices is defined precisely as the limit over the net of finite sets, ordered by inclusion. 2) Order does not matter because the convergence is absolute. You have$$\left\|\sum_{k>N}^M\langle v_k,h\rangle\,v_k\right\|^2=\sum_{k=N+1}^M|\langle v_k,h\rangle|$$ and so Bessel's inequalty, as you mention, implies ... 1 Question was answered in comments by Daniel Fischer: A subset can be open without containing a ball around$0$. A subspace, in the sense of linear subspace, always contains$0$, and if it is open, it therefore contains a ball around$0$. And therefore it is then the entire space. Every infinite-dimensional Hilbert space (that holds for more general ... 4 There exist spaces without a distance measure. What are they good for? Yes, there are things called topological spaces, which don't necessarily have a "distance" function. Even if you don't have a "distance" function, you can still describe the "separation" of points using things called "neighbourhoods". This is a pretty abstract idea, and it ... 2 In general you would use the holomorphic functional calculus $$f(T)=\frac1{2\pi i}\,\int_\Gamma \,f(z)\,(z-T)^{-1}\,dz,$$ where$\Gamma$is a curve in the complex plane with the spectrum of$T$inside the region it delimits, and$f$is an analytic function on that region. Now, if$H$is finite-dimensional, things are way easier. Then you have ... 0 You have the assumption that $$\forall \phi \in \mathcal{H} \quad e_{t}(\phi) = \langle \phi, K(\cdot, t)\rangle _{\mathcal{H}}$$ Consider the meaning of the above statement. The thing on the right implicitly says that$K(\cdot,t)$is an element of$\mathcal H$, because otherwise the inner product is not defined. Any functional of this form is bounded. ... 1 Perhaps the most common type of convergence of metric spaces is Gromov-Hausdorff convergence. However, isometric spaces are distance 0 in the Gromov-Hausdorff metric, so this doesn't work well here. Another idea is convergence of subsets. By extending functions by zero, you can make each of your Hilbert spaces a subset of$L^2(R^n)$. Then a sequence of ... 1 The existence of$J$would imply that$P_1D$and$P_2D$are isometric. This is not the case in general: let$H_1=\mathbb C^2$,$H_2=\mathbb C^3$, and $$D=\{(0,a)\oplus(b,c,0):\ a,b,c\in\mathbb C\}.$$ Then$P_1D$is one-dimensional, while$P_2D$is two-dimensional. 2 a) If$v$is orthogonal to$S$, then it is orthogonal to any linear combination of elements of$S$(this follows from the linearity of the scalar product), i.e. it is orthogonal to the linear span of$S$. b) If$v$is orthogonal to some set$S$, then it is orthogonal to its closure (this follows from the continuity of the scalar product, i.e. if$x_n \to x$... 2 If$H$has infinite dimension, the shift operator under a base is Fredholm with index$-1$. As$ind$is a groups homomorphism between$\text{Fredholms}/\text{Compact}$and$\mathbb{Z}$, and$-1$is a generator of$\mathbb{Z}$, this proves that$ind$is indeed surjective. 1 The first question to answer is: what is your definition of$W^{s,2}(\Omega)$? I will disconsider this question try to help you with your problem. Assume that$\Omega \subset \mathbb{R}^n$. Also, let's suppose that$s\in (0,1)$. Let$u\in L^2(\Omega)$and define $$|u|_{2,s}=\int\int\left(\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\right)^{1/2}dxdy$$ Define ... 1 Yes, it is a Hilbert space. On$\mathbb R^n$this fact is easy to see because the Fourier transform. On other domains, some form of integrated divided difference is used: $$\|f\|^2_{\dot H^s} = \iint_{\Omega\times \Omega} \frac{|f(x)-f(y)|^2}{|x-y|^{n+2s}}\,dx\,dy$$ (The above is the square of a seminorm; add$\int_\Omega |f|^2$to make it the square of a ... 1 Yes, it is possible. Fix a total orthogonal system$(e_i)_i$on your Hilbert space$H$and consider the conjugation of coordinations: $$\varphi:\sum_i\alpha_ie_i \ \mapsto \ \sum_i\bar{\alpha_i}e_i$$ this is an antilinear isomorphism of$H$to itself, so composing it with the Riesz presentation$H^*\to H$gives a linear isomorphism$H^*\to H$. 2 Hint: You can consider how to do each of the following four steps: Step 1: Show that the range of$T_j$and the range of$T_k$(reps. the ranges of$T_j^*$and of$T_k^*$) are orthogonal if$j\ne k$. Step 2: Let$E_k$be the closure of the range of$T_k^*$, and let$F$be the closure of the subspace spanned by all the$E_k$. Show that if$j\ne k$, then ... 1 Yes, the elements of a projective Hilbert space are precisely the one-dimensional subspaces of a linear Hilbert space. The term ray appears to be common in the projective space literature, though some people might associate this word with half-line in a real space. Calling them lines or complex lines would be better, in my opinion. 4 Yes. You should follow Steven Gubkin's advice and draw a picture. After you've done that, you'll understand the intuition behind the following outline of the proof: The continuous function on$[0,1]$can be approximated uniformly by$C^2$functions by the Weierstrass Theorem. So for$f \in L^2[0,1]$you can find$f_1 \in C([0,1])$such that$\|f - f_1\|_2 ...

2

Hint. We can be write $$T = A + iB$$ where $A$ and $B$ are self-adjoint operators defined by $$A := \frac 1 2 (T + T^+)\\ B := -\frac i 2 (T - T^+)$$

2

You can use the sesquilinear forms.Especially a square form. So let's say that $f:H \times H-> \Bbb C$ is a square form that $f(x,x)=<Tx,x>$ then you can so that $\lVert f \rVert \leq \lVert T \rVert \leq 2\lVert f \rVert$. By sesquilinear map we mean a map $φ$:$H\times H->\Bbb C$ with properties: 1)The map is linear to the first variable,the ...

1

To better illustrate the issue, subtract off the weak limits, so that you have two weakly null sequences $u_k$, $v_k$. Suppose they do not converge strongly. Then neither $\|u_k\|$ nor $\|v_k\|$ tend to $0$. But you want to have $\langle u_k,v_k\rangle \to 0$, which means the vectors $u_k$ and $v_k$ must become nearly orthogonal when $k$ is large. There ...

1

There certainly exist references for this topic. Here is the one I found the most helpful. Stone's Theorem in C*-algebras, J. HOLLEVOET, J. QUAEGEBEUR and S. VANKEER I found it in the references of the paper On generalized Stone's Theorem, Massoud Amini which is suggested by user "Argument" in the comments. There is also apparently a book, though I ...

1

First suppose we have a convergent sequence $(Au_n)$ in the range of $A$. We'd like to show that the limit of $(Au_n)$ is also in the range. Note that $(Au_n)$ is Cauchy, and since $\|u_n - u_m\| \leq \|Au_n - Au_m\| /\beta$, we have that $(u_n)$ is Cauchy as well. By completeness $u_n \rightarrow u$ for some $u \in H$. By continuity $Au_n \rightarrow ... 0 An eigenspace of a bounded operator$A$is closed, because it is the kernel of$A-\lambda I$. More generally, an eigenspace of a closed operator$A$is closed (the proof is not hard, but seems tangential to this question – it begins by considering the intersection of the graph of$A$and the graph of$\lambda I$, which is clearly closed, and then projecting ... 2 You have an inner product on$V\otimes W$, so it(s completion) is a Hilbert space; the dual of a Hilbert space is a Hilbert space, by the Riesz representation theorem, which tells you the inner product. 0 You don't need to know that$e^{inx}$form a basis: the completeness part of the basis definition is irrelevant here. The orthogonality of the exponential functions quickly leads to conclusion. Indeed, fix$m\in\mathbb Z$. For every$k\ge |m|$we have $$\int_0^{2\pi} e^{-imx}\sum_{|n|\le k} a_n e^{inx} \,dx = \sum_{|n|\le k}a_n \int_0^{2\pi} e^{i(n-m)x} ... 1 The main idea was given by Daniel Fischer in a comment. It suffices to connect every element f\in L^2(\Omega,\mathbb Z) to zero. Let I=\int_\Omega |f|^2. For 0<t<I pick x=x(t)\in\mathbb R such that \int_{\Omega\cap (-\infty,x)}|f|^2=t. Let f_t=f\chi_{(-\infty,x)} and also f_0=0, f_I=f. Observe that for t<s$$\|f_t-f_s\|_{L^2}^2 ... 2 Hint: try to compute$X^\perp.$2 a) You use that the compacts are an ideal. b) Let$T$be the operator defined on the canonical basis by$Te_{2j-1}=0$,$Te_{2j}=e_{2j+1}$,$j\in\mathbb N$. Then$T$is not compact, as it maps an infinite orthonormal set into another. But$T^2=0$, which of course is compact. c) For any$T$, if$T^*T$is compact, then so is$T$. Because then ... 1 Since this is homework, I am only supposed to give hints.$ A \le B \Leftrightarrow UAU^* \le UBU^* $for any invertible$U$. Hint 1:$ A \le B \Rightarrow B^{-1/2} A B^{-1/2} \le I $. Hint 2: If$V$is invertible, then$ VV^* \le I \Rightarrow V^*VV^*V \le V^* V \Rightarrow V^*V \le I$(use$U = (V^*V)^{-1/2}$for the last implication). Hint 3: So$ ...

4

Look at the image of the unit ball of $L^2$ in $L^1$. If you need more hints, feel free to add a comment.

0

Begin with some orthonormal basis $(e_n)$. Then form another ONB $\mathcal F$ by the following algorithm: $\mathcal F=()$, empty sequence. Find the smallest $n$ such that $e_n$ is not in the span of $\mathcal F$. Add it to the end of $\mathcal F$ and apply Gram-Schmidt. In the finite sequence $\mathcal F=(f_k)$, find the smallest $k$ such that $Tf_k$ is ...

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If $T^*$ is symmetric then we would have that a symmetric linear operator $T$ is ad-joint,because $T^{**}=T$. But this is false in general. Check here when this can be true here

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So, $P$ is a self-adjoint operator such that $P^2=P$. Why $\operatorname{ran}P=\{x:Px=x\}$? Because the second set is contained in the first, tautologically. The first is contained in the second, because if $x=Py$, then $Px=P^2y=Py=x$. Why is $\operatorname{ran}P$ closed? Because it coincides with the kernel of the operator $I-P$, by part 1. Why ...

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Solution №1. Every bounded sequence $\{x_n\}_{n\in\mathbb{N}}$ have weakly convergent subsequence, provided $X^*$ is separable. Let $i:X\to X^{**}$ be the natural embedding into the second dual. Without loss of generality $\{x_n\}_{n\in\mathbb{N}}\subset B_X$. If $X^*$ is separable, then $(B_{X^{**}},w^*)$ is metriazable. By Banach-Alaoglu theorem it ...

0

For any (finite) dimension $d$, $\mathcal{L}^2(\mathbb{R}^d)$ is a separable, infinite dimensional, Hilbert space.

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That matrix is called the Gram-matrix of the vectors $\{ \phi_i \}$.

1

The adjoint is defined by $$T^{*}(x_1, x_2, \ldots, x_{2n}, \ldots ) = (0, x_1, \ldots, 0, x_n 0, x_{n+1} \ldots),$$ that is, $x_k \mapsto 0$ if $k$ is odd, $x_k \mapsto x_{\frac{k}{2}}$ if $k$ is even. Then you can see that $\left<x, T^*y \right> = \sum_{n=1} x_n z_n$, where $z_n = \left\{ \begin{array}{ll}0 &\mbox{ for } n \mbox{ odd} \\ ... 1 The self-adjoint extensions of a symmetric operator is not unique, as the example in question shows. One can define the Cayley transform$C_A$of a symmetric operator$A$.$C_A$is a partial isometry. Self-adjoint extensions of$A$correspond to unitary extensions of$C_A$. It's not hard to see there can be many (or no) unitary extensions for a given ... 1 Another easy non-Hilbert example: Take$X = C_0(\mathbb{R})$with the uniform norm.$X^*$is the Radon measures on$\mathbb{R}$. The sequence$\{ \chi_{[n, n + 1]} \}_n$goes to$0$weakly but is not Cauchy in norm. 3 Consider$\ell^p$, the space of real sequences for which$\sum_j|x_j|^p$is finite with natural norm. If$p\neq 2$, it's not a Hilbert space. If$1\lt p\lt \infty$, take$e_{j}=(0,\dots,0,1,0,\dots)$, where the$1$is at the$j$-th position. This sequence converges weakly to$0$in$\ell^p$but not strongly. The case$p=1$is different. If weak convergence ... 3 The equivalence$(b)\iff (c)$is true in general for "projections"$P, Q$(i.e. idempotency$P^2=P$and$Q^2=Q$suffices). Hint:$PQ=Q \iff (I-P)Q=0$and$N(I-Q)=R(Q)$. The equivalence$(c)\iff (d)$is trivial. Hint: adjoint. It remains to show that$(a)$is equivalent to, say,$(b)$. You were off to a good start. One way to do that, is to prove the ... 2 Yes. Let$Y=X^\perp.$Then for every$y\in Y$we have$\langle x,y\rangle=\lim_{n\to\infty} \langle x_n,y\rangle=0.$Hence$x\in Y^\perp=X.$1 Solved it. :-) I shall post it here on the off chance somebody else ever has the same problem. Conversely, suppose$(T-\lambda)v=0$. We wish to show that$E_{\{\lambda\}}v=v$. This is achieved by proving that the following representation of the spectral projector$E_{\{\lambda\}}$is valid (it is also known as Riesz's formula): If$\gamma\$ is a simple, ...

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